Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes of the charges are doubled and their separation is also doubled, then what happens to the electrical force between

these two charges?
A) It is quadrupled
B) It is halved
C) It is doubled
D) It remains the same

C) It is doubled

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Asked 10/28/2010 1:34:14 PM

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Suppose you pour water into a container until it reaches a depth of 36.7 cm. Next, you carefully pour in a 10.4 cm thickness of olive oil so that it floats on top of the water. What is the pressure at the bottom of the container? Assume 1000 kg/m3 and 940 kg/m3 for the density of water and the oil, respectively.

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Updated 293 days ago|10/4/2013 1:48:34 PM

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Solution:

Formula:

P = Dgh

Where P = Pressure, D is the Density, g = gravity and h = Depth

First we will calculate the pressure due to the oil and this pressure is termed as P1. Similarly, the Density of oil is D1 and Depth of oil is h1.

D1 = 940 kg/m3

h1 = 10.4 cm

g = 9.81 m/s2

P1 = D1*g*h1 = 940 * 9.81 * 10.4 = 95902.56 N/m^2

Now, we will calculate the pressure due to the oil and this pressure is termed as P2. Similarly, the Density of oil is D2 and Depth of oil is h2.

D2= 1000 kg/m3

g = 9.81 m/s2

h2 = 36.7 cm

P2 = D2*g*h2 = 1000 * 9.81 * 36.7 = 360027 N/m^2

The pressure at the bottom of the container is the total pressure = P1 + P2

Thus, P = P1 + P2

= 455929.56 N/m^2

Formula:

P = Dgh

Where P = Pressure, D is the Density, g = gravity and h = Depth

First we will calculate the pressure due to the oil and this pressure is termed as P1. Similarly, the Density of oil is D1 and Depth of oil is h1.

D1 = 940 kg/m3

h1 = 10.4 cm

g = 9.81 m/s2

P1 = D1*g*h1 = 940 * 9.81 * 10.4 = 95902.56 N/m^2

Now, we will calculate the pressure due to the oil and this pressure is termed as P2. Similarly, the Density of oil is D2 and Depth of oil is h2.

D2= 1000 kg/m3

g = 9.81 m/s2

h2 = 36.7 cm

P2 = D2*g*h2 = 1000 * 9.81 * 36.7 = 360027 N/m^2

The pressure at the bottom of the container is the total pressure = P1 + P2

Thus, P = P1 + P2

= 455929.56 N/m^2

Added 293 days ago|10/4/2013 1:48:34 PM

When a glass rod is rubbed with a piece of silk it acquires a positive charge and the silk acquires a negative charge. This process signifies that
A) the glass rod gained electrons from the piece of silk
B) the glass rod lost protons to the piece of silk
C) the glass rod gained protons from a piece of silk
D) the glass rod lost electrons to the piece of silk

Weegy: D) the glass rod lost electrons to the piece of silk
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Two charges, Q1 and Q2, are separated by a certain distance R. If the magnitudes of the charges are doubled and their separation is also doubled, then what happens to the electrical force between these two charges?
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B) It is halved
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